Grade by Grade
A. Exponential and Logarithmic Functions · Exponential and Logarithmic Functions

Switching Between Exponential and Logarithmic Form

A1.3 A1.1 A1.2About 70 minutes
Learning goals
  • I can rewrite an exponential equation in logarithmic form and a logarithmic equation in exponential form.
  • I can solve simple logarithmic equations such as log₄ x = 3 or log_x 49 = 2 by rewriting them in exponential form.
  • I can solve simple exponential equations such as 3ˣ = 20 by rewriting them in logarithmic form.
  • I can isolate the power before switching forms, as in 5(2ˣ) = 60.

Words to know

exponential form
An equation written as bʸ = x.
logarithmic form
The same equation written as log_b x = y.
exponential equation
An equation with the unknown in an exponent, such as 3ˣ = 20.
isolate
Get the power (or the logarithm) by itself on one side of the equation before switching forms.

Two languages, one fact

Think of exponential form and logarithmic form as two languages that say the same thing. Being able to translate quickly between them is the most useful skill in this unit, because whichever form hides the unknown, the other form often sets it free.

bʸ = x ⟺ log_b x = y (b > 0, b ≠ 1, x > 0)

The base stays the base in both. The exponent in one form is the value of the logarithm in the other.

exponenty = 3raise base 4to that power: 4ʸnumberx = 64take log₄log₄ x4³ = 64log₄ 64 = 3undoundo
Raising 4 to the power 3 gives 64; taking log₄ of 64 gets back to 3. Each operation undoes the other.
Worked example

Example 1: Translating

(a) Write 6² = 36 and 10^(−0.5) ≈ 0.3162 in logarithmic form. (b) Write log₂ 128 = 7 and log₅ 0.04 = −2 in exponential form.

  1. (a) Base 6, exponent 2, result 36: log₆ 36 = 2.
  2. (a) Base 10, exponent −0.5, result 0.3162: log 0.3162 ≈ −0.5 (base 10, so no base is written).
  3. (b) Base 2, value 7, argument 128: 2⁷ = 128.
  4. (b) Base 5, value −2, argument 0.04: 5⁻² = 0.04. (Check: 5⁻² = 1/25 = 0.04 ✓)

log₆ 36 = 2; log 0.3162 ≈ −0.5; 2⁷ = 128; 5⁻² = 0.04

Solving simple logarithmic equations

If the unknown is the argument or the base of a logarithm, rewrite in exponential form. Then the unknown is no longer trapped inside a log.

Worked example

Example 2: Unknown argument or unknown base

Solve (a) log₃ x = 4, (b) log₄ x = 3/2, (c) log_x 49 = 2, (d) log_x 8 = 3/2.

  1. (a) Exponential form: x = 3⁴ = 81.
  2. (b) x = 4^(3/2) = (√4)³ = 2³ = 8.
  3. (c) x² = 49, so x = 7 or x = −7. A base must be positive, so reject −7. x = 7.
  4. (d) x^(3/2) = 8. Raise both sides to the power 2/3 (the reciprocal of 3/2): x = 8^(2/3) = (∛8)² = 2² = 4.
  5. Check (d): log₄ 8 = 3/2 because 4^(3/2) = 8 ✓

(a) 81 (b) 8 (c) 7 (d) 4

Always check the restrictions

After solving, make sure the base is positive and not 1, and the argument is positive. Answers that break these rules must be rejected, as x = −7 was in Example 2(c).

Solving simple exponential equations

If the unknown is in the exponent, rewrite in logarithmic form. Then use the change of base formula from lesson 2 to get a decimal.

First isolate the power. In 5(2ˣ) = 60 the power is 2ˣ, so divide by 5 before you switch forms. Do not divide 5(2ˣ) into "10ˣ": 5 × 2ˣ is not 10ˣ.

Worked example

Example 3: Unknown exponent

Solve to three decimal places: (a) 3ˣ = 20, (b) 10ˣ = 45, (c) 5(2ˣ) = 60, (d) 2^(x − 1) = 10.

  1. (a) Logarithmic form: x = log₃ 20 = log 20 ÷ log 3 ≈ 2.727.
  2. (b) x = log 45 ≈ 1.653.
  3. (c) Isolate the power: 2ˣ = 12. Then x = log₂ 12 = log 12 ÷ log 2 ≈ 3.585.
  4. (d) The exponent is the whole expression x − 1: x − 1 = log₂ 10 ≈ 3.322, so x ≈ 4.322.
  5. Check (c): 5(2^3.585) ≈ 5(12.00) = 60 ✓

(a) x ≈ 2.727 (b) x ≈ 1.653 (c) x ≈ 3.585 (d) x ≈ 4.322

Worked example

Example 4: A growing culture

A lab culture starts with 200 bacteria and doubles every 3 hours, so N = 200(2)^(t/3) after t hours. When will there be 5000 bacteria?

  1. Set N = 5000: 200(2)^(t/3) = 5000.
  2. Isolate the power: 2^(t/3) = 25.
  3. Logarithmic form: t/3 = log₂ 25 = log 25 ÷ log 2 ≈ 4.644.
  4. Multiply by 3: t ≈ 13.93.
  5. Reasonable? After 12 h (4 doublings) there are 200 × 16 = 3200; after 15 h, 6400. 5000 is between them ✓

After about 13.9 hours.

t (h)N2468101214161000200030004000500060007000(0, 200)t ≈ 13.9 hN = 200(2)^(t/3)N = 5000
The horizontal line N = 5000 meets the growth curve at t ≈ 13.9 h, the solution of 200(2)^(t/3) = 5000.
Just for fun

Riddle: Why did the exponent never worry about getting lost?

Because it knew the logarithm would always find it.

Hands-on activity

Log memory game

You need: 16 index cards (or squares of paper) and a pen. A family member can play too.

  1. On 8 cards, write exponential facts: 2⁵ = 32, 3⁴ = 81, 10⁻² = 0.01, 4^(1/2) = 2, 5⁰ = 1, 7² = 49, 8^(2/3) = 4, (1/3)⁻² = 9.
  2. On the other 8 cards, write the same facts in logarithmic form, for example log₂ 32 = 5.
  3. Shuffle and lay all 16 face down. Take turns turning over two cards. If they say the same fact, keep them and say both forms aloud; otherwise turn them back.
  4. Harder round: write only the logarithmic side with a blank (log₂ 32 = ?) and the matching card shows only the answer.

What to look for: matching pairs are log₂ 32 = 5, log₃ 81 = 4, log 0.01 = −2, log₄ 2 = 1/2, log₅ 1 = 0, log₇ 49 = 2, log₈ 4 = 2/3 and log_(1/3) 9 = −2. Watch for players who swap the 5 and 32: the value of the logarithm is always the exponent.

Your turn

1

Write 2⁻³ = 1/8 in logarithmic form.

2

Write log₉ 3 = 1/2 in exponential form.

3

Solve log₂ x = 6.

4

Solve log_x 125 = 3.

5

Solve 4ˣ = 30 to three decimal places.

6

Solve 3(5ˣ) = 450 to three decimal places.

7

Solve 10^(x + 2) = 7 to three decimal places.

More practice: warm-up

1

Write 10⁴ = 10 000 in logarithmic form.

2

Write log₃ 9 = 2 in exponential form.

3

Solve log₅ x = 2.

4

Solve log x = 3.

5

Solve 2ˣ = 16.

More practice: on your own

1

Solve log₈ x = −2/3.

2

Solve log_x (1/16) = −2.

3

Solve 7^(2x) = 300 to three decimal places.

4

Solve 2(3ˣ) − 5 = 31 to three decimal places.

5

A forest of 4000 trees grows by 5% a year, so T = 4000(1.05)ᵗ. In how many years will there be 6000 trees?

Challenge

1

Solve log₂ (log₃ x) = 2.

2

Solve log_(x − 1) 36 = 2.

3

Solve 5^(x²) = 40 exactly and to three decimal places.

Check your understanding

Quiz

1.

Which is the logarithmic form of 3⁵ = 243?

2.

Which is the exponential form of log₆ x = 2?

3.

Solve log₂ x = −4.

4.

Solve log_x 81 = 4.

5.

What is the first step in solving 6(4ˣ) = 96?

6.

To three decimal places, the solution of 2ˣ = 50 is

For parents and teachers

Big idea (A1.3): bʸ = x and log_b x = y are the same statement. When the unknown is the argument or base of a logarithm, switch to exponential form; when the unknown is an exponent, switch to logarithmic form.

How to help: Ask "Where is the unknown?" before your teen starts. If it is inside a logarithm, go to exponential form; if it is in an exponent, go to log form. Ask them to check every answer by substituting it back.

Common difficulties:

  • Combining a coefficient with the base: 5(2ˣ) is not 10ˣ. Divide by 5 first.
  • Forgetting that the base must be positive (rejecting x = −7 in log_x 49 = 2).
  • In 2^(x − 1) = 10, forgetting that the whole x − 1 is the exponent.
  • Rounding the logarithm before multiplying or adding.